Question

5. Determinaţi valorile numerelor raționale a şi b, pentru care: a(5+2+√3)+b(4-3√3) = 6+7√3.​

Answer 1

Răspuns:

$a = -\frac{10}{23},\, b = \frac{47}{23}$

Rezolvare:

$a(5+2\sqrt{3})+b(4-3\sqrt{3}) = 6+7\sqrt{3} \Rightarrow$

$\Rightarrow (a\cdot 5+a\cdot 2\sqrt{3})+(b\cdot 4-b\cdot 3\sqrt{3}) = 6+7\sqrt{3}$

$\Rightarrow (5a+2a\sqrt 3)+(4b-3b\sqrt{3}) = 6+7\sqrt{3}$

$\Rightarrow (5a+4b)-(2a\sqrt{3}-3b\sqrt{3}) = 6+7\sqrt{3}$

$\Rightarrow (5a+4b)-\sqrt{3}(2a-3b) = 6+7\sqrt{3}$

$\Rightarrow (5a+4b)-(2a-3b)\sqrt{3} = 6+7\sqrt{3}$

$\Rightarrow (5a+4b)+[-(2a-3b)\sqrt{3}] = 6+7\sqrt{3}$

$\Rightarrow (5a+4b)+(-2a+3b)\sqrt{3} = 6+7\sqrt{3}$

$\Rightarrow \left.\begin{cases} 5a+4b = 6|\cdot 2 \\ -2a+3b = 7|\cdot 5 \end{cases}\right| \Rightarrow \left.\begin{cases} 10a+8b = 12 \\ -10a+15b = 35 \end{cases}\right|(+) \Rightarrow$

$\Rightarrow 10a-10a+8b+15b = 12+35 \Rightarrow 23b = 47 \Rightarrow \boxed{b = \dfrac{47}{23}}$

$5a+4b = 6 \Rightarrow 5a+4\cdot \dfrac{47}{23} = 6\Big|\cdot 23 \Rightarrow 115a+4\cdot 47 = 138\Rightarrow$

$\Rightarrow 115a+188 = 138 \Rightarrow 115a = 138-188 \Rightarrow 115a = -50 \Rightarrow$

$\Rightarrow a = \dfrac{-50}{115} \Rightarrow \boxed{a = -\dfrac{10}{23}}$